873. Length of Longest Fibonacci Subsequence

Description

A sequence x1, x2, ..., xn is Fibonacci-like if:

• n >= 3
• xi + xi+1 == xi+2 for all i + 2 <= n

Given a strictly increasing array arr of positive integers forming a sequence, return the length of the longest Fibonacci-like subsequence of arr. If one does not exist, return 0.

A subsequence is derived from another sequence arr by deleting any number of elements (including none) from arr, without changing the order of the remaining elements. For example, [3, 5, 8] is a subsequence of [3, 4, 5, 6, 7, 8].

 

Example 1:

Input: arr = [1,2,3,4,5,6,7,8]
Output: 5
Explanation: The longest subsequence that is fibonacci-like: [1,2,3,5,8].

Example 2:

Input: arr = [1,3,7,11,12,14,18]
Output: 3
Explanation: The longest subsequence that is fibonacci-like: [1,11,12], [3,11,14] or [7,11,18].

 

Constraints:

• 3 <= arr.length <= 1000
• 1 <= arr[i] < arr[i + 1] <= 109

Solutions

Solution 1: Dynamic Programming

We define \(f[i][j]\) as the length of the longest Fibonacci-like subsequence, with \(\textit{arr}[i]\) as the last element and \(\textit{arr}[j]\) as the second to last element. Initially, for any \(i \in [0, n)\) and \(j \in [0, i)\), we have \(f[i][j] = 2\). All other elements are \(0\).

We use a hash table \(d\) to record the indices of each element in the array \(\textit{arr}\).

Then, we can enumerate \(\textit{arr}[i]\) and \(\textit{arr}[j]\), where \(i \in [2, n)\) and \(j \in [1, i)\). Suppose the currently enumerated elements are \(\textit{arr}[i]\) and \(\textit{arr}[j]\), we can obtain \(\textit{arr}[i] - \textit{arr}[j]\), denoted as \(t\). If \(t\) is in the array \(\textit{arr}\), and the index \(k\) of \(t\) satisfies \(k < j\), then we can get a Fibonacci-like subsequence with \(\textit{arr}[j]\) and \(\textit{arr}[i]\) as the last two elements, and its length is \(f[i][j] = \max(f[i][j], f[j][k] + 1)\). We can continuously update the value of \(f[i][j]\) in this way, and then update the answer.

After the enumeration ends, return the answer.

The time complexity is \(O(n^2)\), and the space complexity is \(O(n^2)\), where \(n\) is the length of the array \(\textit{arr}\).

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def lenLongestFibSubseq(self, arr: List[int]) -> int:
        n = len(arr)
        f = [[0] * n for _ in range(n)]
        d = {x: i for i, x in enumerate(arr)}
        for i in range(n):
            for j in range(i):
                f[i][j] = 2
        ans = 0
        for i in range(2, n):
            for j in range(1, i):
                t = arr[i] - arr[j]
                if t in d and (k := d[t]) < j:
                    f[i][j] = max(f[i][j], f[j][k] + 1)
                    ans = max(ans, f[i][j])
        return ans

class Solution {
    public int lenLongestFibSubseq(int[] arr) {
        int n = arr.length;
        int[][] f = new int[n][n];
        Map<Integer, Integer> d = new HashMap<>();
        for (int i = 0; i < n; ++i) {
            d.put(arr[i], i);
            for (int j = 0; j < i; ++j) {
                f[i][j] = 2;
            }
        }
        int ans = 0;
        for (int i = 2; i < n; ++i) {
            for (int j = 1; j < i; ++j) {
                int t = arr[i] - arr[j];
                Integer k = d.get(t);
                if (k != null && k < j) {
                    f[i][j] = Math.max(f[i][j], f[j][k] + 1);
                    ans = Math.max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    int lenLongestFibSubseq(vector<int>& arr) {
        int n = arr.size();
        int f[n][n];
        memset(f, 0, sizeof(f));
        unordered_map<int, int> d;
        for (int i = 0; i < n; ++i) {
            d[arr[i]] = i;
            for (int j = 0; j < i; ++j) {
                f[i][j] = 2;
            }
        }

        int ans = 0;
        for (int i = 2; i < n; ++i) {
            for (int j = 1; j < i; ++j) {
                int t = arr[i] - arr[j];
                auto it = d.find(t);
                if (it != d.end() && it->second < j) {
                    int k = it->second;
                    f[i][j] = max(f[i][j], f[j][k] + 1);
                    ans = max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
};

func lenLongestFibSubseq(arr []int) (ans int) {
    n := len(arr)
    f := make([][]int, n)
    for i := range f {
        f[i] = make([]int, n)
    }

    d := make(map[int]int)
    for i, x := range arr {
        d[x] = i
        for j := 0; j < i; j++ {
            f[i][j] = 2
        }
    }

    for i := 2; i < n; i++ {
        for j := 1; j < i; j++ {
            t := arr[i] - arr[j]
            if k, ok := d[t]; ok && k < j {
                f[i][j] = max(f[i][j], f[j][k]+1)
                ans = max(ans, f[i][j])
            }
        }
    }

    return
}

function lenLongestFibSubseq(arr: number[]): number {
    const n = arr.length;
    const f: number[][] = Array.from({ length: n }, () => Array(n).fill(0));
    const d: Map<number, number> = new Map();
    for (let i = 0; i < n; ++i) {
        d.set(arr[i], i);
        for (let j = 0; j < i; ++j) {
            f[i][j] = 2;
        }
    }
    let ans = 0;
    for (let i = 2; i < n; ++i) {
        for (let j = 1; j < i; ++j) {
            const t = arr[i] - arr[j];
            const k = d.get(t);
            if (k !== undefined && k < j) {
                f[i][j] = Math.max(f[i][j], f[j][k] + 1);
                ans = Math.max(ans, f[i][j]);
            }
        }
    }
    return ans;
}

use std::collections::HashMap;
impl Solution {
    pub fn len_longest_fib_subseq(arr: Vec<i32>) -> i32 {
        let n = arr.len();
        let mut f = vec![vec![0; n]; n];
        let mut d = HashMap::new();
        for i in 0..n {
            d.insert(arr[i], i);
            for j in 0..i {
                f[i][j] = 2;
            }
        }
        let mut ans = 0;
        for i in 2..n {
            for j in 1..i {
                let t = arr[i] - arr[j];
                if let Some(&k) = d.get(&t) {
                    if k < j {
                        f[i][j] = f[i][j].max(f[j][k] + 1);
                        ans = ans.max(f[i][j]);
                    }
                }
            }
        }
        ans
    }
}

/**
 * @param {number[]} arr
 * @return {number}
 */
var lenLongestFibSubseq = function (arr) {
    const n = arr.length;
    const f = Array.from({ length: n }, () => Array(n).fill(0));
    const d = new Map();
    for (let i = 0; i < n; ++i) {
        d.set(arr[i], i);
        for (let j = 0; j < i; ++j) {
            f[i][j] = 2;
        }
    }
    let ans = 0;
    for (let i = 2; i < n; ++i) {
        for (let j = 1; j < i; ++j) {
            const t = arr[i] - arr[j];
            const k = d.get(t);
            if (k !== undefined && k < j) {
                f[i][j] = Math.max(f[i][j], f[j][k] + 1);
                ans = Math.max(ans, f[i][j]);
            }
        }
    }
    return ans;
};

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